# [LeetCode][39. Combination Sum] Easy Backtracking Approach: Deduplicating and Pruning

*By Long Luo*

This article is the solution Easy Backtracking Approach: Deduplicating and Pruning of Problem 39. Combination Sum.

# Backtracking

It’s a typical **Backtracking** algorithm.

We can easily draw the tree below.

Take \(\textit{target} = 7\) as the root node and subtract the value of the edge while creating a child node.

Stop it when the value of the node is \(0\) or a negative number.

All paths from the root node to leaf node \(0\) are the results of the question.

Let’s coding.

1 | class Solution { |

Take \(\textit{target} = 7\) for example, \([[2, 2, 3], [2, 3, 2], [3, 2, 2], [7]]\) are the answer.

However, the correct answer is \([[7], [2, 2, 3]]\).

Since it requires that each solution is not **in order**, so the \([2, 2, 3]\) and \([2, 3, 2]\) are the same answer.

What’s problem?

When and how did we **duplicate** the path?

# Why Duplicated?

At each node of the tree, **all the candidates** were searched and used, so there is a duplicate list.

Is it possible to **deduplicate** while searching?

Of course we can.

We have to **search the elements in order** while searching, since the output is not by order. We search from the certain position in the next round.

1 | public static void backtrack(List<List<Integer>> res, List<Integer> list, int[] candidates, int start, int remain) { |

By now we **deduplicating** it and complete the problem.

# Pruning

We can prune the tree to make it work more efficiency.

1 | class Solution { |

## Analysis

**Time Complexity**: \(O(Sum)\), \(Sum\) is all the path.**Space Complexity**: \(O(\textit{target})\).

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